Two batteries with emf epsilon_1 and epsilon_2, with internal resistances r_1 an

Question

Two batteries with emf epsilon_1 and epsilon_2, with internal resistances r_1 and r_2 respectively, are connected as shown in the diagram below. (Assume epsilon_1 = 12 v and r_1 = 1 ohm.) Calculate the magnitude and indicate the direction of flow of current in the figure shown above. Epsilon_2 = 28.0 v and r_2 = 0.60 ohm. magnitude What is the net voltage when the two batteries are connected in opposition to each other? What is the net internal resistance of this circuit? A direction Find the terminal voltage of each battery. What is the sign of the current if it is flowing into the positive terminal of the battery? V What is the sign of the current if it is flowing into the positive terminal of the battery? V

Explanation / Answer

Given Data

E1 = 12 V

E2 = 28 V

r1 = 1 ohm

r2 = 0.6 ohm

Solution :-

a) Aplly KVL to the circuit considering anticlock wise

-0.6*I + 28 -12 -1*I = 0

-1.6 I + 16 = 0

-1.6 I = - 16

I = 10 A

b) V1 = E1 + I*r1

V1 = 12 + 10*1

V1 = 22 V

V2 = E2 - I*r2

V2 = 28 - 10*0.6

V2 = 22 V

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